Spearman Rank Correlation Coefficient

8/3/2019 Spearman Rank Correlation Coefficient

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Statistical Methodand Advanced Application


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SPEARMAN RANK

CORRELATION COEFFICIENT

SMALL SAMPLELARGE SAMPLE

APPROXIMATION


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SMALL SAMPLE


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Spearman Rank Correlation

Coefficient

Learning Outcome

Student should be able to make

a relationship between two

variable by using Spearman rankcorrelation coefficient.


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Assumptions

a)The data consist of a random sample ofn

pairs of numeric or non-numeric

observations.

b)Each pair of observations represents two

measurements taken on the same object orindividual, called the unit of association.


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Procedures

1)If the data consist of observations from a bivariate population, wedesignate the n pairs of observations (X1,Y1), (X2,Y2),, (Xn,Yn).

2)EachXis ranked relative to all other observed values ofX, from

smallest to largest in order of magnitude. The rank of the ith value of

Xis denoted by R(Xi )=1 ifXi is the smallest observed value ofX.

3) Each Yis ranked relative to all other observed values ofY, from

smallest to largest in order of magnitude. The rank of the ith value of

Yis denoted by R(Yi )=1 ifYi is the smallest observed value ofY.

4)If ties occur among theXs or among the Ys, each tied value is

assigned the mean of the rank positions for which it is tied.

5)If the data consist of nonnumeric observations, they must be capable

of being ranked as described.


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Hypotheses

Case A

(two-sided)

Case B

(one-sided)

Case C

(one-sided)

H0 : X and Y are

independent.H0 : X and Y are

independent.

H0 : X and Y are

independent.

H1 : X and Y are

either directly

or inversely

related.

H1 : There is a

direct

relationship

between X andY.

H1 : There is an

inverse

relationship

between Xand Y.

= 0

0

= 0 = 0sH :0 sH :0 sH :0

sH :1 0:H s1 > 0:H s1 <


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Test Statistic

1) Rank all sample observations from smallest to

largest.

2)Find di and di2where di=R(Xi)R(Yi), and di is

the difference in the ranks given to the variablesXi and Yi.

3)Sum the square of the difference in the ranks

observations from populationXi and Yi, di2

4)The test statistic is,)1(

61

2

2

nn

dr

i

s


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5) rsis to measure the strength of the relationship

between the sampleXand Yvalues and as an

estimate of the strength of the relationship

betweenXand Yin the sampled population.

6) When the rank ofXis the same as the rank ofY

for every pair of observations (perfect direct

relationship), all the differences di will be equalto zero, and rs will be equal to +1.

7) Kendall(T3)* has shown that in general rs = -1

when the rank of one variable within each pair of

observations (Xi, Yi) is the reverse of the other

(perfect inverse relationship).

*Kendall, M.G.,Rank Correlation Methods, fourth edition, London: Griffin, 1970.


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Thus if

[R(X) = 1, R(Y) = n]

[R(X) = 2, R(Y) = n-1],.,[R(X) = n, R(Y) = 1]

for n pairs of observations, rs = -1. This may be illustrated

by means of a simple example. Suppose have the followingpairs of observations of (Xi, Yi) : (0,10),(8,3),(2,9),(5,6).

The ranks are

So, the di2=(-3) 2 +(3) 2 + (-1) 2 + (1) 2 = 20. Then

rs = 1[6(20)/4(16-1) = -1

8) Kendall also shows that rs can never be greater than +1 or

less than -1.

Xi 0 8 2 5

R(Xi) 1 4 2 3

Yi 10 3 9 6

R(Yi) 4 1 3 2


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Decision Rule

Case Reject H0 if

A

(Two-sided)

B

(One-sided positive)

C

(One-sided negative)

2rr

s

rr

s

rr

s

2rr

s


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Example 9.1(pg 360)

Pincherle and Robinson(E1)* note a marked interobservervariation in blood pressure. They found that doctors whoread high on systolic tended to read high on diastolic.Table 9.1 shows mean systolic and diastolic bloodpressure readings by 14 doctors. We wish to compute a

measure of the strength of the relationship between thetwo variables. Under the consumption that these 14doctors constitute a random sample from a population ofdoctors, we wish to know whether we may conclude fromthe data that there is a direct relationship between systolicand diastolic readings. Suppose we let = 0.05

*G.Pincherle and D.Robinson, Mean Blood Pressure and its Relation to Other Factors

Determined at a Routine Executive Health Examination, J.Choronic Dis.,27 (1974),245-260;used with permission of Pergamon Press.


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Mean blood pressure readings, millimeters mercury, by doctor

Doctor Systolic Diastolic

1 141.8 89.72 140.2 74.4

3 131.8 83.5

4 132.5 77.8

5 135.7 85.8

6 141.2 86.5

7 143.9 89.4

8 140.2 89.3

9 140.8 88.0

10 131.7 82.2

11 130.8 84.6

12 135.6 84.4

13 143.6 86.3

14 133.2 85.9


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1) HYPOTHESES

Systolic and diastolic blood pressure readings by

doctors are independent.

There is a direct relationship between systolic and

diastolic blood pressure readings by doctors

(claim).

:0H

:1H

2) TEST STATISTIC


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2) TEST STATISTICSYSTOLIC

( )

DIASTOLIC

( )R( ) R( ) = R ( )R( )

141.8 89.7 12 14 - 2 4

140.2 74.4 8.5 1 7.5 56.25131.8 83.5 3 4 -1 1

132.5 77.8 4 2 2 4

135.7 85.8 7 7 0 0

141.2 86.5 11 10 1 1

143.9 89.4 14 13 1 1

140.2 89.3 8.5 12 - 3.5 12.25

140.8 88.0 10 11 -1 1

131.7 82.2 2 3 -1 1

130.8 84.6 1 6 -5 25

135.8 84.4 6 5 1 1

143.6 86.3 13 9 4 16

133.2 85.9 5 8 -3 9

iX

iX iYi

Y ix iYid2id

50.132=2id


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By using the formula of test static:

From the table, we get the value ofand n = 14. Then, substitute the values into the

equation above.

50.1322

id

)1-14(14

)50.132(6-12

sr

7088.0

)1(

6-1

2

2

nn

dr is


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3) CRITICAL VALUE

Table A.21 reveals that, for n = 14 and (1) = 0.05,

the critical value of is 0.464.s

r


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4) DECISION

Since 0.7088 > 0.464, we reject .

5) CONCLUSION

There is enough evidence to support the claim

that the doctors who read high on systolic

tend to read high on diastolic blood pressure.

0H


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Exercise 1

Table 9.4 shows the serum and bone magnesium levels of 14 patients are reported by

Alfrey et al.*

. Can we conclude from these data that a relationship exist betweenserum magnesium and bone magnesium in the sample population?

Serum Mg ( m Eq./L.) Bone Mg ( m Eq./kg ash )3.60 6722.85 6102.80 6212.70 5672.60 5702.55 6382.55 6122.45 5522.25 5241.80 4001.45 2771.35 2941.40 3380.90 230

*Alfrey, Allen C.,Nancy L. Miller, and Donald Butkus, Evaluation Of Body Magnesium

Stores,J.Lab. Clin. Med.,84 (1974), 153-1


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Exercise 2

Ten seventh-grade children randomly selected from a certain public school

system were ranked according to the quality of their home environment and thequality of their performance in school. The result are shown in table 9.45.

compute rs and determine whether one can conclude that the two variable are

directly related.

Ten seventh-grade children ranked according to quality of home

environment and quality of performance in school.

Child Home environment Performance in school1 3 12 7 93 10 84 9 105 2 36 1 47 6 58 4 29 8 610 5 7


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Exercise 3

The Spearman's Rank Correlation Coefficient is used to discover the strength of a link

between two sets of data. This example looks at the strength of the link between the

price of a convenience item (a 50cl bottle of water) and distance from the

Contemporary Art Museum (CAM ) in El Raval, Barcelona. compute rs and

determine whether one can conclude that the two variable are inversly related.

Convinence store Distance from CAM ( m) Price of 50cl bottle ()1 50 1.802 175 1.203 270 2.004 375 1.005 425 1.006 580 1.207 710 0.808 790 0.609 890 1.00

10

980

0.85


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Large-Sample Approximation

When the sample size is greater than 100, we

cannot use Table A.21 to test the significance

ofrs. Then we may compute

which is distributed approximately as thestandard normal.

1 nrzs


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LINK OF YOUTUBE

http://www.youtube.com/watch?v=Eu_XOoFNfR4&feature=mfu_in_order&list=UL
http://www.youtube.com/watch?v=Eu_XOoFNfR4&feature=mfu_in_order&list=ULhttp://www.youtube.com/watch?v=Eu_XOoFNfR4&feature=mfu_in_order&list=ULhttp://www.youtube.com/watch?v=Eu_XOoFNfR4&feature=mfu_in_order&list=ULhttp://www.youtube.com/watch?v=Eu_XOoFNfR4&feature=mfu_in_order&list=UL


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Spearman Rank Correlation Coefficient